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LM2734: switch current limit (protection) vs actual output load current

Carl Ungvarsky
Prodigy 130 points
Part Number: LM2734
Hi,
I'm analyzing a good working LM2734X circuit. I'm looking for advice on how to use the data given in the data sheet to predict the behavior when a fault happens that triggers the internal protection system. I'm also interested in the behavior when a fault condition is just below the threshold of the internal protection system.
My aim is to determine the maximum input current and output current that can flow in the case of a circuit fault. The kind of circuit faults I have in mind are things such as a short circuit or a near-short circuit in the load, or in the inductor or capacitor(s) at the voltage regulator output.
The LM2734 data sheet (rev K) description says it can drive 1 ampere loads (text in sections 1 and 3). Nowhere in the specifications (section 6) is a max Iout given. The "switch current limit" Icl is given as 1.2A to 2.5A, with 1.7A typical under certain conditions. This switch current is that of the internal NMOS switch.
Since I can control the load current through varying the load, but I cannot directly control the switch current, I figure I need to increase the load current to increase the internal switch current. When the switch current reaches the current limit threshold (somewhere in the 1.2A to 2.5A range) the output will be disabled. From that, I will know the load current that led to the current limit shutdown, but I still won't know the corresponding switch current value.
Question 1: Is there a better way to determine the internal switch current limit under my operating conditions?
Question 2: Can we predict the current limit given our specific operating conditions? (Figures 3 and 4 in the data sheet don't match my operating conditions)
Question 3: Is there a calculation or approximation for internal switch current given the load current?
Regards
Carl
  • 0
    TI__Guru**** 191445 points

    Carl,

    First off, the output is not disabled during an over current.  LM2734 uses cycle by cycle current limit, so the high side FET is turned off when the current (which is ramping up during the on time) reaches the current limit point.  It turns on again at the next cycle.  The net effect is that the duty cycle is limited to less than what is required for regulation so the output voltage drops.  The more the load exceeds the limit, the lower the voltage will drop.

    You can calculate the internal switch current:

    Ipk = Iload + 1/2 * ((vout/Vin) * (Vin - Vout) / (Fsw * Lout))

    You cannot predict the exact current limit point from a calculation.  It will be somewhere in between 1.2 A and 1.5 A.  The best practice if you need need the full 1 A load current is to limit your p-p inductor (equal to (vout/Vin) * (Vin - Vout) / (Fsw * Lout)) current to less than 400 mA so that the equation above equals 1.2 A.

    Also if you plan on operating into overload conditions, you should choose your inductor so that the saturation current is above 2.5 A.

    Let me know if you have further questions.

  • 0 in reply to JohnTucker
    Prodigy 130 points

    Hi John.

    In my use case, the current draw will be well below 1 A. What I'm looking into here is the behavior in a fault condition to verify that the current limit works as advertised. I also want to check that when this protection mechanism works, overheating or burning of my external circuit parts is prevented.

    Thanks

    Carl

  • 0 in reply to Carl Ungvarsky
    TI__Guru**** 191445 points

    Carl,

    If you want to check current limit functionality here is how We do it.

    You will need to lift the non-switching end of the output inductor and add a wire loop back to the PCB pad long enough to accommodate a clamp on current probe.  You can monitor the output voltage, switch node voltage, and inductor current.  Increase the load current and observe the peak inductor current, which is the same as the switch current reach the current limit point.  Increasing the load current beyond this point will cause the output voltage to decrease.  Eventually, the output will approach O V, put the output current will still be at Icl - ILp-p/2.

    Does this clarify your question?